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Proof step 21 from proof hassharing-validrefs-dups (previous step)

validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), path∃(str, r, r, st), ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ hasSharing(r, st)

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
validrefs(str, refs, ty, st), refs ≠ [] ⊦ ∃ sks. fits(refs .first, sks, st) ∧ 1 + # sks = # refs ∧ (∀ n. 0 < n ∧ n ≤ # sks → st[refs .first -- firstn(n, sks)] .refval = refs[n])
Proof step 20:
∃ sks. fits((r ' + refs1 + r ') .first, sks, st) ∧ 1 + # sks = # (r ' + refs1 + r ') ∧ (∀ n. 0 < n ∧ n ≤ # sks → st[(r ' + refs1 + r ') .first -- firstn(n, sks)] .refval = (r ' + refs1 + r ')[n]), validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), path∃(str, r, r, st), ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ hasSharing(r, st)

Rule argument

[LEMMAARG : ex-validrefs-fits-stronger, [str, refs, ty, st] → [str, r ' + refs1 + r ', st[r -- _type] .type, st]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])], [EXTRATERMS : (str, r ' + refs1 + r ', st[r -- _type] .type, st)]]